USAPhO 2021, Полуфинал — Задача 3. A3 · Electroneering

• a. $\mu = \frac{1}{3}qcR$
• b. $R = \frac{3}{2}\frac{\hbar}{mc}$
• c. $U_E = \frac{1}{3}\alpha E_0$
• d. $U_B \sim \alpha E_0$
• e. Both U_E and U_B are much smaller than E₀, by a factor of α ≪ 1, so this classical model cannot explain the origin of the electron’s rest energy.

a.

A point on the sphere’s equator moves at a speed $\omega R$, where $\omega$ is the angular velocity of rotation. Setting $\omega R = c$ gives $\omega = c/R$.

The spinning sphere can be thought of as a stack of infinitesimal current loops, all of which have a magnetic moment pointing in the same direction. Consider making a thin, circular slice of the sphere’s surface, corresponding to polar angles in the range $(\theta, \theta + d\theta)$. This slice has a radius $R\sin\theta$, so that the surface area of the slice is

The charge of the slice is

Since the charge $dQ$ moves around the rotation axis one time per period $T = 2\pi/\omega$, the corresponding current is

The magnitude of the magnetic moment of this slice is

Using the provided integral, the total magnetic moment is

If you weren’t able to do this, you could also have given the answer $\mu \sim qcR$, which can be derived by dimensional analysis, for partial credit.

Alternative solution: Note that for a uniformly charged ring of mass $dm$, charge $dq$, and radius $r$, rotating with angular velocity $\omega$, the ratio of the magnetic moment and the angular momentum is

The ratio is independent of $r$ and $\omega$. Since the sphere can be decomposed into such rings, the total magnetic moment and total angular momentum must have the same ratio,

Finally, we know that $L = (2/3)mR^2\omega$ for a spherical shell. Plugging this in and using $\omega = c/R$ gives $\mu = qcR/3$ as before, but with no integration required.

b.

Since the magnetic moment is fixed, and we want the radius to be small, we want the electron to be spinning as fast as possible. Thus, the magnetic moment has the value found in part (a), and equating this to the known value gives

Again, you can get $R \sim \hbar/mc$ by dimensional analysis.

c.

For a collection of charges, the total energy stored in the electrostatic field is

where $V_i$ is the electric potential at $q_i$. In this case, the total charge is $q$, and all of the charge is at potential $q/4\pi\epsilon_0R$, so

Using the result of part (b),

Note that you can’t get this answer by dimensional analysis alone, since $\alpha$ is dimensionless. (However, if you found $R$ by dimensional analysis, and additionally reasoned that $U_E$ could depend only on $q$, $\epsilon_0$, and $R$, then you could derive $U_E \sim \alpha E_0$, for partial credit.)

d.

Following the hint, we can estimate

where $B_0$ is the typical magnetic field inside the sphere. Actually finding the value of $B_0$ would require doing some complicated integrals. To get a rough estimate, note that if we replaced the sphere with a ring of charge, then at the center of the ring,

Thus, we have

To get this in terms of the fine structure constant, we use $c^2 = 1/\mu_0\epsilon_0$, giving

An even faster way to get this result is to note that in general, the energy stored in magnetic fields tends to be a factor of order $(v/c)^2$ smaller than the energy stored in electric fields, where $v$ is the speed of the charge. In this problem the charge is all moving relativistically, so we must have $U_B \sim U_E$.

e.

Both $U_E$ and $U_B$ are much smaller than $E_0$, by a factor of $\alpha \ll 1$. Thus, this classical model cannot explain the origin of the electron’s rest energy.

There were many attempts to make classical models of the electron in the early 20$^{\mathrm{th}}$ century, but they all ran into difficulties like this one. For more on this subject, see chapter II-28 of the Feynman lectures.